Polynomial and Rational Inequalities Questions with Solutions

Explore a variety of multiple-choice questions focused on polynomial and rational inequalities, complete with detailed solutions and step-by-step explanations to help deepen your understanding. If you're new to the topic or want a refresher, consider starting with this comprehensive tutorial on inequalities before tackling the problems.

Question 1

Solve the inequality:

\[ -(x + 2) + 2x > 2(x - 3) + 3x \]
  1. \(\text{No solutions}\)
  2. \((-\infty, \dfrac{4}{5})\)
  3. \((\dfrac{4}{5}, +\infty)\)
  4. \((-\infty, +\infty)\)
  5. \((-\infty, 1)\)
View Solution and Explanation
  • Given inequality: \[-(x + 2) + 2x > 2(x - 3) + 3x\]
  • Expand both sides:
    Left side: \(-(x + 2) + 2x = -x - 2 + 2x = x - 2\)
    Right side: \(2(x - 3) + 3x = 2x - 6 + 3x = 5x - 6\)
  • Simplified inequality: \[x - 2 > 5x - 6\]
  • Move variables and constants: \[x - 5x - 2 > -6 \implies -4x - 2 > -6 \implies -4x > -4\]
  • Divide both sides by \(-4\) and reverse the inequality sign: \[x < \dfrac{-4}{-4} = 1\]
  • Final answer in interval notation: \[\boxed{(-\infty, 1)}\]

Question 2

Solve the polynomial inequality:

\[ (6x + 2) (10 - 20x) \ge 0 \]
  1. \( \left( -\dfrac{1}{3}, \dfrac{1}{2} \right) \)
  2. \( \left[ -\dfrac{1}{2}, +\infty \right) \)
  3. \( \text{No solutions} \)
  4. \( \left( -\infty, 0 \right] \)
  5. \( \left[ -\dfrac{1}{3}, \dfrac{1}{2} \right] \)
View Solution and Explanation
  • Find the roots (zeros) of the expression by setting each factor to zero: \[6x + 2 = 0 \implies x = -\dfrac{1}{3}\] \[10 - 20x = 0 \implies x = \dfrac{1}{2}\]
  • The expression changes sign at the critical points \(x = -\dfrac{1}{3}\) and \(x = \dfrac{1}{2}\).
  • Test interval signs:
    • \(x < -\dfrac{1}{3}\) (e.g., \(x = -1\)): \((-6+2)(10+20) = -120 < 0\)
    • \(-\dfrac{1}{3} < x < \dfrac{1}{2}\) (e.g., \(x = 0\)): \((2)(10) = 20 > 0\)
    • \(x > \dfrac{1}{2}\) (e.g., \(x = 1\)): \((8)(-10) = -80 < 0\)
  • Include endpoints since the inequality includes \(\ge\): \[x \in \left[-\dfrac{1}{3}, \dfrac{1}{2}\right]\]
  • Final answer in interval notation: \[\boxed{\left[-\dfrac{1}{3}, \dfrac{1}{2}\right]}\]

Question 3

Solve the inequality:

\[ x^2 + 3x - 4 > 0 \]
  1. \( (-4, 1) \)
  2. \( (-\infty, -4) \cup (1, +\infty) \)
  3. \( (1, +\infty) \)
  4. \( \text{No solutions} \)
  5. \( (-\infty, -4] \cup [1, +\infty) \)
View Solution and Explanation
  • Factor the quadratic expression: \[x^2 + 3x - 4 = (x + 4)(x - 1) > 0\]
  • Find critical points: \(x = -4\) and \(x = 1\).
  • Test intervals:
    • \((-\infty, -4)\) (e.g., \(x = -5\)): \((-1)(-6) = 6 > 0\)
    • \((-4, 1)\) (e.g., \(x = 0\)): \((4)(-1) = -4 < 0\)
    • \((1, \infty)\) (e.g., \(x = 2\)): \((6)(1) = 6 > 0\)
  • Final answer in interval notation: \[\boxed{(-\infty, -4) \cup (1, \infty)}\]

Question 4

Solve the inequality:

\[ 2x^2 < 32 \]
  1. \( (-4,\, 4) \)
  2. \( (-\infty,\, 4) \)
  3. \( \left(-\sqrt{32},\, \sqrt{32} \right) \)
  4. No solutions
  5. \( (-4,\, +\infty) \)
View Solution and Explanation
  • Divide both sides by 2: \[x^2 < 16\]
  • Take square roots considering both positive and negative roots: \[-4 < x < 4\]
  • Final answer in interval notation: \[\boxed{(-4,\ 4)}\]

Question 5

Solve the inequality:

\[ 25 x^2 \le 9x \]
  1. No solutions
  2. \( (-\infty , +\infty) \)
  3. \( [0 , +\infty) \)
  4. \( \left[0 , \dfrac{9}{25} \right] \)
  5. \( \left[0 , \dfrac{9}{5} \right) \)
View Solution and Explanation
  • Rearrange into standard form and factor: \[25x^2 - 9x \le 0 \implies x(25x - 9) \le 0\]
  • Critical points: \(x = 0\) and \(x = \dfrac{9}{25}\).
  • Test intervals to find where the expression is less than or equal to zero, yielding: \[\left[0, \dfrac{9}{25}\right]\]
  • Final answer in interval notation: \[\boxed{\left[0, \dfrac{9}{25}\right]}\]

Question 6

Solve the inequality:

\[ \dfrac{1 + x} {|x - 2|} \gt 0 \]
  1. \( (-1 , +\infty) \)
  2. \( (-1 , 2) \cup (2 , +\infty) \)
  3. \( [-1 , +\infty) \)
  4. \( (2 , +\infty) \)
  5. \( (-\infty , +\infty) \)
View Solution and Explanation
  • Note that the denominator \(|x - 2|\) is always positive except at \(x = 2\).
  • Thus, the sign depends entirely on the numerator \(1 + x > 0 \implies x > -1\), with the domain restriction \(x \neq 2\).
  • Final answer in interval notation: \[\boxed{(-1, 2) \cup (2, \infty)}\]

Question 7

Solve the inequality:

\[ \dfrac{(2x^{2} + 3)(x + 2)}{x^{2} - 4} \geq 0 \]
  1. \((-\infty, +\infty)\)
  2. \((-\infty, 2)\)
  3. \((2, +\infty)\)
  4. \([2, +\infty)\)
  5. \( \text{no solutions} \)
View Solution and Explanation
  • Factor the denominator: \[\dfrac{(2x^2 + 3)(x + 2)}{(x - 2)(x + 2)} \geq 0\]
  • Cancel common factors with the restriction \(x \neq -2\): \[\dfrac{2x^2 + 3}{x - 2} \geq 0\]
  • Since the numerator \(2x^2 + 3\) is always positive, the sign depends solely on \(x - 2 > 0 \implies x > 2\).
  • Final answer in interval notation: \[\boxed{(2, \infty)}\]

Question 8

Solve the inequality:

\[ \dfrac{4}{6x - 4} \leq \dfrac{2}{2x + 2} \]
  1. \((-\infty, -1] \cup [4, +\infty)\)
  2. \(\left(\dfrac{2}{3}, 4\right]\)
  3. \((-1, \dfrac{2}{3})\)
  4. \((-1, \dfrac{2}{3}) \cup [4, +\infty)\)
  5. \([4, +\infty)\)
View Solution and Explanation
  • Simplify and combine terms into a single rational expression: \[\dfrac{-x + 4}{(3x - 2)(x + 1)} \leq 0\]
  • Critical points are \(x = -1\), \(x = \dfrac{2}{3}\), and \(x = 4\).
  • Testing intervals reveals the solution region including zero in the numerator and excluding denominator zeros.
  • Final answer in interval notation: \[\boxed{(-1, \dfrac{2}{3}) \cup [4, \infty)}\]

Question 9

Solve the double inequality:

\[ 2 < \dfrac{|x + 1|}{2} - 2 < 5 \]
  1. \((-3, -15)\)
  2. no solutions
  3. \((-15, -9) \cup (7, 13)\)
  4. \((7, 13)\)
  5. \((-\infty, +\infty)\)
View Solution and Explanation
  • Isolate the absolute value expression: \[4 < \dfrac{|x + 1|}{2} < 7 \implies 8 < |x + 1| < 14\]
  • Solve compound inequalities for absolute value to get intervals \(-15 < x < -9\) and \(7 < x < 13\).
  • Final answer in interval notation: \[\boxed{(-15, -9) \cup (7, 13)}\]

Question 10

Solve the inequality:

\[ x^{2} < -x^{4} \]
  1. no solutions
  2. \((-1, 1)\)
  3. \((-\infty, +\infty)\)
  4. \((0, +\infty)\)
  5. \([0, +\infty)\)
View Solution and Explanation
  • Rewrite in standard form: \[x^2 + x^4 < 0 \implies x^2(1 + x^2) < 0\]
  • Since \(x^2 \geq 0\) and \(1 + x^2 > 0\) for all real numbers \(x\), their product can never be less than zero.
  • Final answer: \[\boxed{\text{No real solutions}}\]

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